Area under a Curve and between Curves

Transcript

0:00Introduction

Hello everyone! Okay, a small puzzle before we start. Take the curve y equals x squared, minus four x, plus three, and integrate it from one to three. You get minus four thirds. So the area under this curve is minus four thirds? Can an area really be negative? Something is wrong here, right? Today: the area under a curve, and the area between two curves. These questions come up often in JEE, and they are very scoring. So, let's begin.

0:33Area under a curve is never negative

So, first, the region. On screen: the curve y equals f of x, the x axis, and the vertical lines x equals a and x equals b. Okay? The long S is the integral sign, a and b are its limits, and d x means we add thin vertical strips along x. So the area is just the integral from a to b of f of x, d x, right? Almost, but not quite, because an area can never be negative. So we take the modulus: the two straight bars around f of x, read as mod f of x. Area is the integral from a to b of the modulus of f of x, d x. Clear? Now think. Where the curve goes below the x axis, f of x is negative. So what does the plain integral give there? Yes, exactly, that part comes out negative! The less-than sign says f of x is below zero there, so every thin strip has a negative height. And when the curve is partly above and partly below, the negative part cancels some of the positive part, so the plain integral gives less than the true area. Right? So find where f changes sign, that is, where the curve cuts the x axis. Split there, take every piece as positive, and add. Got it?

1:55Between two curves: upper minus lower

Now, two curves, and the region trapped between them. Take a thin vertical strip. Its height is y upper minus y lower; the small words under each y name the curve. Add all the strips: the integral from a to b of the bracket, y upper minus y lower, d x. Okay? But usually nobody gives you a and b. So where do they come from? From the intersection points, correct! Put the two equations equal and solve. Where the curves meet, there are your limits. With me? And always sketch first: the sketch tells you which curve is upper. Upper minus lower, never the other way round. And if the picture is symmetric, find one half and double it. Clear?

2:42Horizontal strips

Sometimes vertical strips are painful, so we turn them sideways. Each horizontal strip has length x right minus x left; the small words under each x name the right and left curves. So the area is the integral from c to d of x right minus x left, d y, where c and d are the lowest and highest y values. Okay? So, when should you choose horizontal strips? Perfect, when the curves are easier to write as x equals g of y! Like parabolas lying on their side. Right? Here is one from the course: x equals y squared, and x equals two minus y squared. Put them equal: y squared is one, so y is plus or minus one. The plus-minus sign means both. Got it? The right curve is two minus y squared, the left is y squared, so the length is two minus two y squared. Integrate from minus one to one, d y, and the area is eight thirds. Clean and simple, with no square roots at all!

3:49Three standard areas to remember

Now, three standard areas that save you minutes. First, the ellipse: x squared over a squared, plus y squared over b squared, equals one, with semi-axes a and b. Its area is pi, the Greek letter on screen, times a times b. Okay? Second, between the parabolas y squared equals four a x, opening to the right, and x squared equals four a y, opening upwards. The area is sixteen a squared over three. Right? Third, the parabola y squared equals four a x, cut off by its latus rectum: the vertical line x equals a, through the focus. That area is eight a squared over three.

4:34Example 1: y = x2 - 4x + 3 below the x-axis, area 4/3

Clear? Now, remember our puzzle from the start? Here it is, from the course. Find the area bounded by y equals x squared minus four x plus three, and the x axis. Pause the video and try it first. Done? Okay, let's check. Factorise: x squared minus four x plus three is x minus one, times x minus three. So the curve cuts the axis at x equals one and x equals three. These are our limits. Right? Now, between one and three, is the curve above the axis, or below it? Test one point. At x equals two: four minus eight plus three, minus one. So y is less than zero. So the whole arc from one to three lies below the axis. Clear? Now the plain integral. The antiderivative is x cubed over three, minus two x squared, plus three x. The square brackets, with one below and three on top, mean: value at three, minus value at one. At three: nine minus eighteen plus nine, zero. At one: one third minus two plus three, four thirds. So we get minus four thirds. Okay? So the area is minus four thirds, right? No, no, careful here, that is exactly the trap from our puzzle! Area can never be negative. The curve is below the axis, so take minus the integral, which is its modulus. The area is four thirds square units, and our puzzle is solved!

6:11Example 2: area between y = mod x and y = 2 - x2 is 7/3

Now a harder one, also from the course. Find the area enclosed by y equals mod x and y equals two minus x squared. In the picture, the modulus of x is the V shape, and the parabola opens downwards. The answer, seven thirds, is even written inside. Your job is to prove it. Pause the video and try it first. Done? Okay, let's check. The picture is symmetric about the y axis, so take x greater than or equal to zero, where the modulus of x is just x. Put the curves equal: two minus x squared equals x. With me? So x squared plus x minus two equals zero. The double arrow means, so it follows: x plus two, times x minus one, equals zero. Okay? So the curves meet at x equals one, and at x equals minus two, right? Oh no, careful here: we took x greater than or equal to zero, so minus two is not allowed. Reject it. So they meet at x equals one, and by symmetry at minus one: the two dots in the picture. Clear? The upper curve is the parabola, the lower one is y equals x. So the area is two times the integral from zero to one of two minus x squared minus x, d x. The two in front covers the left half. Right? From zero to one, the three terms give two, one third and one half. Two minus one third minus one half is seven sixths. Times two: seven thirds. That is exactly the number in the picture, very good!

7:52Quick revision: mod f, upper minus lower, horizontal strips

So, quick revision. Area is the integral of the modulus of f of x: never negative, so split where the curve crosses the axis. Between curves: upper minus lower, limits from the meeting points. For curves like x equals g of y: right minus left, integrate in y. And always sketch first, and use symmetry. Now go and try the practice problems on this page. All the best!